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: Can someone check my working on these Diff Eqs.

  1. Aug 15, 2006 #1
    URGENT: Can someone check my working on these Diff Eqs.

    Hi I have a maths exam tomorrow, and want to make sure Im doing the right thing, please bare in mind i study physics, and am not used to some of the techniques that are used in pure maths, so if the person can help using the methods i am using, would be much appreciated.


    Question 1)
    Solve this first order differential equation

    dy/dx = x^3/y

    subject to the boundary condition that y(x=0)=1


    ok so what I do is rearrange to get:

    1/y dy = x^3 dx

    and then integrate both sides with respect to their variable to get

    ln y = 1/4 x^4 + c

    so then i remove the ln:

    y = e^(1/4 x^4 + c)

    SO
    now im a little bit confused with the boundary conditions, am I right in saying that when x = 0 , y = 1??

    if that is true then am I right in saying that the constant of integration must be 0 (as e^0 = 1), so the answer is:

    y = e ^ (1/4 x^4)

    ??

    Question 2)
    Solve the 2nd order differential equation:

    d^2y/dx^2 - 4dy/dx = 0

    (no boundary conditions this time)

    ok so here i used the substitution U = dy/dx and get

    dU/dx - 4U = 0

    and rearrange to get:

    1/U dU = 4 dx

    and similarly to above integrate both sides

    ln U = 4x + c1

    rearrange:

    dy/dx = U = e ^ (4x +c1)

    ok, so now im slightly confused, do I ignore the constant becasue there are no bc's in this question???? If that is right, then I just integrate again to get:

    y = 4 e^(4x) + C

    is that correct??
     
  2. jcsd
  3. Aug 15, 2006 #2

    benorin

    User Avatar
    Homework Helper

    Question #1 is correct: well-done.
    Question #2 is correct, except that you need the constant of integration to remain, so your answer is [tex]y=\frac{1}{4}e^{4x+C_1}+C_2[/tex] and it should be 1/4 not 4. This may be simplified since [tex]e^{4x+C_1}=e^{4x}e^{C_1}=Ce^{4x},\, C>0[/tex]
     
  4. Aug 16, 2006 #3
    wow thanks for the quick reply, unfortunately I wasnt able to access the site to check for a reply yesterday..

    question 1 is wrong though, look closely, the LHS isn't 1/y.... Just realised myself, the exam went well anyways.


    THANKSSSS!!!!!!
     
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